Changes

Contestant 2 will never expend effort greater than <math>v_2\,\</math>, and contestant 1 can win for sure is he expends an epsilon more, so this constraints the effort of both parties to lie on the range <math>[0,v_2]\,\</math>. If contestant 1 expends an epsilon more than <math>v_2\,\</math> he gaurantees a payoff of <math>v_1 - v_2\,\</math>; this is his equilibrium payoff. If he does this then player 2's payoff is zero; this is his equilibrium payoff! Under mixed strategies a player must be indifferent at all points, we use this indifference to solve the game:

:<math>\pi_1(x_1) = F_2(x_1)v_1 - x_1 = v_1-v_2\,\</math>

:<math>\pi_2(x_2) = F_1(x_2)v_2 - x_2 = 0\,\</math>

Together this can be used to solve for <math>F_1\,\</math> and <math>F_2\,\</math>, yielding the results above. Using the standard expected value formula (and differentiating the CDFs to get PDFs first):

:<math>\mathbb{E}(X) = \int_{-\infty}^\infty x f(x)\, \dx\,\</math>

<center><math>\mathbb{E}x_1 = \frac{v_2}{2} \,</math> and <math>\mathbb{E}x_2=\frac{v_2^2}{2v_1}\,\</math></center>